Checking if mySql datetime is older then 1 day from php now() - php

Question

I have a record returned from MySQL that has a datetime field. What I want to do is take this value and see if it is older then 24 hours, I presume using PHP's time() to get the current time.
At the moment if I echo them out I get:
1276954824 this is php's time()
2010-06-19 09:39:23 this is the MySQL datetime
I presume the top one is a unix time? Have been playing around with strtotime but with not much success..
ANy help welcome!

Answer 1

No success?
echo strtotime("2010-06-19 09:39:23");
gives me
1276940363
(mktime(9, 39, 23, 6, 19, 2010) gives the same time, so the parsing works correctly)
To get the differences in seconds, you can substract the timestamps, e.g.
$diff = time() - strtotime("2010-06-19 09:39:23");
If the differences is larger than 86400 (60*60*24) seconds, then the timestamps are more than one day apart:
if(time() - strtotime("2010-06-19 09:39:23") > 60*60*24) {
// timestamp is older than one day
}

Answer 2

You can also do:
SELECT * FROM my_table WHERE timestamp < NOW() - INTERVAL 1 DAY;

Answer 3

Why are you mixing PHP times and MySQL times?
Instead, do the comparison directly in MySQL:
To get the current date/time in MySQL use the NOW() function. You can compare, for example, 2010-06-19 09:39:23' < DATE_SUB(NOW(), INTERVAL 1 DAY)
This would check to see if the given date (presumably in a column) is older than 24 hours.
If it's absolutely necessary to convert a MySQL timestamp to a UNIX timestamp, you can use MySQL's UNIX_TIMESTAMP() function to do so.

Answer 4

I wrote a function, by which you can determine if the first given date is one day or n days bigger or smaller than the second given date.
$date1 = "2013/03/01";
$date2 = "2013/03/01";
$sign = "-";
$diff = 1;
$result = isDaysSmallerBigger($date1, $date2, $sign, $diff);
var_dump($result);
/**
* Note: this function is only supported in php 5.3 or above
* 1. If $date1 - $date2 = $sign $diff days, return true;
* 2. If $date1 equals $date2 and $diff euqals 0, whether $sign is "+" or "-", return true
* 3. Else return false;
* #param unknown_type $date1
* #param unknown_type $date2
* #param string $sign: "-": $date1 < $date2; "+": $date1 > $date2;
* Besides if $diff is 0, then both "-" and "+" means $date1 === $date2;
* #param integer $diff: date difference between two given dates
*/
function isDaysSmallerBigger($date1, $date2, $sign, $diff) {
$dateTime1 = new DateTime($date1);
$dateTime2 = new DateTime($date2);
$interval = $dateTime2->diff($dateTime1);
$dayDiff = $interval->format('%a');
$diffSign = $interval->format('%R');
if((int)$dayDiff === (int)$diff) {
// Correct date difference
if((int)$diff !== 0) {
// Day difference is not 0
if($diffSign === $sign) {
return true;
} else {
return false;
}
} else if((int)$diff === 0) {
// Day difference is 0, then both given "+" and "-" return true
return true;
}
} else {
// Incorrect date difference
return false;
}
}